Question Number 30259 by mondodotto@gmail.com last updated on 19/Feb/18
Commented by prof Abdo imad last updated on 20/Feb/18
$${we}\:{have}\:\:\frac{{x}}{{x}−\mathrm{1}}\:−\frac{\mathrm{1}}{{lnx}}=\:\frac{{xlnx}\:−{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right){lnx}}=\frac{{u}\left({x}\right)}{{v}\left({x}\right)} \\ $$$${we}\:{hsve}\:{u}\left(\mathrm{1}\right)={v}\left(\mathrm{1}\right)=\mathrm{0}\:{let}\:{use}\:{hospital}\:{theorem} \\ $$$${u}^{'} \left({x}\right)=\mathrm{1}+{lnx}\:−\mathrm{1}\Rightarrow{u}^{''} \left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:{anf}\:{u}^{''} \left(\mathrm{1}\right)=\mathrm{1} \\ $$$${v}^{'} \left({x}\right)={lnx}\:+\frac{{x}−\mathrm{1}}{{x}}={lnx}\:+\mathrm{1}\:−\frac{\mathrm{1}}{{x}}\Rightarrow{v}^{''} \left({x}\right)=\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{v}^{''} \left(\mathrm{1}\right)=\mathrm{2}\:\:\Rightarrow\:{lim}\:\left(….\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:. \\ $$
Answered by $@ty@m last updated on 19/Feb/18
![We have ln x=ln {1+(x−1)} =(x−1)−(((x−1)^2 )/2)+... ∴lim_(x→1) ((x/(x−1))−(1/(ln x))) ∴lim_(x→1) ((x/(x−1))−(1/((x−1)−(((x−1)^2 )/2)+(((x−1)^3 )/3)−...))) lim_(x→1) (1/(x−1)){x−(1/(1−((x−1)/2)+(((x−1)^2 )/3)−..))} lim_(x→1) (1/(x−1))[((x{1−((x−1)/2)+(((x−1)^2 )/3)−..}−1)/(1−((x−1)/2)+(((x−1)^2 )/3)−..))] lim_(x→1) (1/(x−1))[(((x−1)−((x(x−1))/2)+((x(x−1)^2 )/3)−..)/(1−((x−1)/2)+(((x−1)^2 )/3)−..))] lim_(x→1) ((x−1)/(x−1))[((1−(x/2)+((x(x−1))/3)−..)/(1−((x−1)/2)+(((x−1)^2 )/3)−..))] =(1/2)](https://www.tinkutara.com/question/Q30296.png)
$${We}\:{have} \\ $$$$\mathrm{ln}\:{x}=\mathrm{ln}\:\left\{\mathrm{1}+\left({x}−\mathrm{1}\right)\right\} \\ $$$$=\left({x}−\mathrm{1}\right)−\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}+… \\ $$$$\therefore\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{{x}}{{x}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{ln}\:{x}}\right) \\ $$$$\therefore\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{{x}}{{x}−\mathrm{1}}−\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)−\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{3}}−…}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}−\mathrm{1}}\left\{{x}−\frac{\mathrm{1}}{\mathrm{1}−\frac{{x}−\mathrm{1}}{\mathrm{2}}+\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{3}}−..}\right\} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}−\mathrm{1}}\left[\frac{{x}\left\{\mathrm{1}−\frac{{x}−\mathrm{1}}{\mathrm{2}}+\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{3}}−..\right\}−\mathrm{1}}{\mathrm{1}−\frac{{x}−\mathrm{1}}{\mathrm{2}}+\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{3}}−..}\right] \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}−\mathrm{1}}\left[\frac{\left({x}−\mathrm{1}\right)−\frac{{x}\left({x}−\mathrm{1}\right)}{\mathrm{2}}+\frac{{x}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{3}}−..}{\mathrm{1}−\frac{{x}−\mathrm{1}}{\mathrm{2}}+\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{3}}−..}\right] \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}−\mathrm{1}}{{x}−\mathrm{1}}\left[\frac{\mathrm{1}−\frac{{x}}{\mathrm{2}}+\frac{{x}\left({x}−\mathrm{1}\right)}{\mathrm{3}}−..}{\mathrm{1}−\frac{{x}−\mathrm{1}}{\mathrm{2}}+\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{3}}−..}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$